Divide the following complex numbers. $ \dfrac{-20+20i}{4+4i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${4-4i}$ $ \dfrac{-20+20i}{4+4i} = \dfrac{-20+20i}{4+4i} \cdot \dfrac{{4-4i}}{{4-4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-20+20i) \cdot (4-4i)} {(4+4i) \cdot (4-4i)} = \dfrac{(-20+20i) \cdot (4-4i)} {4^2 - (4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-20+20i) \cdot (4-4i)} {(4)^2 - (4i)^2} = $ $ \dfrac{(-20+20i) \cdot (4-4i)} {16 + 16} = $ $ \dfrac{(-20+20i) \cdot (4-4i)} {32} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-20+20i}) \cdot ({4-4i})} {32} = $ $ \dfrac{{-20} \cdot {4} + {20} \cdot {4 i} + {-20} \cdot {-4 i} + {20} \cdot {-4 i^2}} {32} $ Evaluate each product of two numbers. $ \dfrac{-80 + 80i + 80i - 80 i^2} {32} $ Finally, simplify the fraction. $ \dfrac{-80 + 80i + 80i + 80} {32} = \dfrac{0 + 160i} {32} = 5i $